3.2 Permutations, Combinations & Probability Distributions

Permutations, Combinations & Probability Distributions

Combinatorics and probability are central to competitive engineering entrance examinations. Candidates must solve intricate counting problems, evaluate constrained arrangements, and compute exact probabilities of compound events swiftly without relying on long exhaustive listings.

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1. Fundamental Principles of Counting

All combinatorial mathematics rests on two foundational logic rules:

The Multiplication Principle (Rule of Product)

If an operation can be performed in $m$ different ways, and following this operation a second independent operation can be performed in $n$ different ways, then the two operations in succession can be performed in $m \times n$ different ways.

Generalization: If a composite task consists of $k$ sequential stages with $n_1, n_2, \dots, n_k$ choices at each stage, the total number of composite outcomes is:

$N = n_1 \times n_2 \times n_3 \times \dots \times n_k$

The Addition Principle (Rule of Sum)

If a task can be performed by executing either alternative $A$ (which can occur in $m$ ways) or alternative $B$ (which can occur in $n$ ways), and alternatives $A$ and $B$ are mutually exclusive (cannot occur simultaneously), then the task can be performed in $m + n$ different ways.

Factorial Algebra & Properties

For any non-negative integer $n$:

$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$

Key conventions and identities:

• Base case: $0! = 1$ (by definition, representing the one way to arrange an empty set).
• Recurrence: $n! = n \times (n-1)!$.
• Undefined values: Factorials of negative integers are undefined (they approach infinity in gamma function analysis).
• Algebraic cancellation: $\frac{(n+2)!}{n!} = (n+2)(n+1) = n^2 + 3n + 2$.

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2. Permutations (Arrangements Where Order Matters)

A permutation is an ordered arrangement of elements chosen from a given set.

Linear Permutation of Distinct Objects

The number of arrangements of $n$ distinct objects taken $r$ at a time ($0 \le r \le n$) is denoted by ${}^n P_r$ or $P(n, r)$:

${}^n P_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\dots(n-r+1)$

• Arranging all $n$ objects linearly: ${}^n P_n = \frac{n!}{(n-n)!} = \frac{n!}{0!} = n!$.
• Choosing 0 objects: ${}^n P_0 = \frac{n!}{n!} = 1$.
• Choosing 1 object: ${}^n P_1 = n$.

Permutations with Repetition (Non-Distinct Elements)

If a collection of $n$ objects contains $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, $\dots$, and $n_k$ identical objects of type $k$ (such that $n_1 + n_2 + \dots + n_k = n$), the number of distinct linear permutations is:

$P = \frac{n!}{n_1! \, n_2! \, n_3! \dots n_k!}$

Classic Problem: How many distinct arrangements can be formed from the letters of the word PAKISTAN?

• Total letters $n = 8$.
• Counts: A appears 2 times, and P, K, I, S, T, N appear 1 time each.
• Total permutations: $\frac{8!}{2! \, 1! \, 1! \, 1! \, 1! \, 1! \, 1!} = \frac{40,320}{2} = 20,160$.

Circular Permutations

When $n$ distinct objects are arranged in a closed loop or circle, shifting all objects uniformly does not alter their relative cyclic positions. Fixing one reference object eliminates the rotational symmetry:

$P_{\text{circle}} = (n-1)!$

Symmetric / Reversible Circular Arrangements (Necklaces & Keyrings): If clockwise and counter-clockwise orders cannot be distinguished (such as turning a necklace or keyring over):

$P_{\text{necklace}} = \frac{(n-1)!}{2}$

Constrained Arrangement Tactics

1. The Bundle (Tie) Method (Items Must Be Together): Treat the constrained elements as a single composite unit (bundle). Permute the resulting set of items including the bundle, then multiply by the internal permutations of the items within the bundle. Example: Arranging 4 boys and 3 girls such that the 3 girls sit together: treat 3 girls as 1 block. There are $4 + 1 = 5$ units to arrange ($5! = 120$ ways). The 3 girls can be internally arranged in $3! = 6$ ways. Total arrangements $= 120 \times 6 = 720$.
2. The Gap Method (Items Must Never Be Adjacent): First arrange the unconstrained items. Then insert the constrained items into the available spaces (gaps) formed before, between, and after the unconstrained items. Example: Arranging 5 men and 3 women so that no two women sit together: first seat the 5 men ($5! = 120$ ways). This creates 6 available gaps (including ends). Choose and arrange 3 women in these 6 gaps in ${}^6 P_3 = 6 \times 5 \times 4 = 120$ ways. Total arrangements $= 120 \times 120 = 14,400$.

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3. Combinations (Selections Where Order Does Not Matter)

A combination is an unordered collection or selection of $r$ objects chosen from a set of $n$ distinct objects:

${}^n C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Connection Between Permutations and Combinations

Because every selection of $r$ items can be internally arranged in $r!$ ways:

${}^n P_r = r! \times {}^n C_r \implies {}^n C_r = \frac{{}^n P_r}{r!}$

Fundamental Properties & Identities of Combinations

1. Symmetry Property: ${}^n C_r = {}^n C_{n-r}$
2. Equal Index Theorem (High-Yield Test Property): $\text{If } {}^n C_x = {}^n C_y, \quad \text{then either } x = y \quad \text{or } x + y = n$
3. Pascal's Addition Identity: ${}^n C_r + {}^n C_{r-1} = {}^{n+1} C_r$
4. Sum of All Combinations: $\sum_{r=0}^{n} {}^n C_r = {}^n C_0 + {}^n C_1 + {}^n C_2 + \dots + {}^n C_n = 2^n$
5. At Least One Item Selected: $\sum_{r=1}^{n} {}^n C_r = 2^n - 1$

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4. Probability Theory Fundamentals

Probability quantifies the likelihood of occurrence of a random event within an experiment.

Sample Space & Events

• Sample Space ($S$): The set of all possible elementary outcomes of a random experiment. For rolling a fair 6-sided die, $S = \{1, 2, 3, 4, 5, 6\}$ with $n(S) = 6$.
• Event ($E$): Any subset of the sample space ($E \subseteq S$).
• Equally Likely Outcomes: Outcomes that have identical physical propensity of occurring.

Classical (Laplacian) Definition of Probability

If an experiment has $n(S)$ mutually exclusive, collectively exhaustive, and equally likely outcomes, and $n(E)$ of them are favorable to event $E$:

$P(E) = \frac{n(E)}{n(S)}$

Probability Axioms:

1. $0 \le P(E) \le 1$ for any event $E$.
2. $P(S) = 1$ (certain event), and $P(\emptyset) = 0$ (impossible event).
3. Complementary Event Rule: $P(E') = 1 - P(E)$.

Speed Technique: Whenever a problem asks for the probability of "at least one" occurrence, immediately compute $1 - P(\text{none})$.

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5. Compound Events & Addition Theorems

Mutually Exclusive (Disjoint) Events

Events $A$ and $B$ are mutually exclusive if they cannot occur at the same time ($A \cap B = \emptyset$):

$P(A \cup B) = P(A) + P(B)$

General Addition Theorem (Non-Mutually Exclusive Events)

If events $A$ and $B$ can occur simultaneously ($A \cap B \ne \emptyset$):

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For three arbitrary events $A, B, C$:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$

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6. Conditional Probability & Independence

Conditional Probability

The probability of event $A$ occurring given that event $B$ has already occurred ($P(B) > 0$) is defined by restricting the sample space to $B$:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

Multiplication Theorem of Probability

Rearranging the conditional probability equation:

$P(A \cap B) = P(B) \times P(A|B) = P(A) \times P(B|A)$

Independent Events

Two events $A$ and $B$ are statistically independent if the occurrence of one provides no information about the probability of the other: $P(A|B) = P(A)$. Consequently:

$P(A \cap B) = P(A) \times P(B) \iff A \text{ and } B \text{ are independent}$

Critical Distinction on the Exam: Never confuse mutually exclusive events with independent events:

• If $A$ and $B$ are mutually exclusive and have non-zero probabilities, $P(A \cap B) = 0 \ne P(A)P(B)$. Therefore, mutually exclusive events with positive probability are dependent.

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7. High-Yield ECAT Problem Archetypes

1. Playing Card Mechanics (52-Card Deck)

• Total cards $= 52$, split into 4 suits of 13 cards each: Spades (♠) and Clubs (♣) are Black (26 cards); Hearts (♥) and Diamonds (♦) are Red (26 cards).
• Each suit contains: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.
• Face cards (Court cards): Jack, Queen, King across 4 suits $= 3 \times 4 = 12$ cards.
• Honor cards: Aces + Face cards $= 16$ cards.

2. Rolling Two Fair Dice

• Total outcomes $n(S) = 6 \times 6 = 36$.
• Sums range from 2 to 12. The number of favorable outcomes follows a triangular distribution centered at sum 7:

3. Sampling: With vs. Without Replacement

• With Replacement: Trials are independent; probabilities remain identical at each stage.
• Without Replacement: Trials are dependent; the denominator and numerator adjust after each selection (hypergeometric model).